I think it would be nice to derive the fictitious forces at least once and then use the definitions later on if you remember.
In this case in a rotating frame of reference the force is given as:
$$
\frac{\vec F}{m} =
\left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right)
\left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right)
\vec x \\
\vec{F} = m
\left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right)
\left( \vec v + \vec \omega \times \vec x \right) \\
=
m \vec a + m \vec \omega \times \vec v
+ m \vec \omega \times \vec v + m\vec \omega \times \vec \omega \times \vec x \\
= m \vec a + m \vec \omega \times \vec \omega \times \vec x + 2 m \vec \omega \times \vec v
$$
Where $\vec a$ is the acceleration in the non rotating frame of reference and $\vec v$ is velocity in the non rotating frame, whereas $\vec F$ is the experienced force in the rotating frame.
From the expression above you can distinguish the centrifugal force term and the Coriolis force term.
Just to ease our interpretation, let's use the vector tripple product formula to reexpress the centrifugal for term:
$$
\vec \omega \times \vec \omega \times \vec x
= \vec \omega \cdot \left( \vec \omega \cdot \vec x \right) -
\vec x \cdot \left| \vec \omega \right|^2
$$
Now the motion of the bug is radial and on the disc, which means that in the centrifugal force term will become:
$$
\vec x \cdot \left| \vec \omega \right|^2
$$
This means that the centrifugal force indeed acts towards the 'outside' of the disc and the Coriolis force will be acting perpendicular to the direction of motion. The friction force will be acting antiparallel to the direction of motion.
Hope that this long and mathematically rigorous way, will let you picture everything better.
Note: The final expression of the fictitious forces can be expressed in the following way as well:
$$
m \vec a = F - m \vec \omega \times \vec \omega \times \vec x - 2 m \vec \omega \times \vec v
$$
This can be interpreted that we are always interested in the acceleration in the non-rotating frame $\left(\vec a \right)$ and that we need to subtract from the force $\left( \vec F \right)$ all the fictitious forces we know about.
So this is not inconsistent with the Coriolis force definition given by joshphysics:
$$
m \vec a = \vec F + \vec F_{centrifugal} + \vec F_{Coriolis}
$$